In order to balance this equation you need to count each element and how many of the individual elements are in the equation.
_H2+N2=2 NH3
You multiply the 2 (Which is the coefficient) by the 3 (which is the subscript) This would equal 6 which indicated there are 6 hydrogen atoms on the right side so the left side should also have 6 hydrogen atoms
The missing coefficient on the left side must multiple the 2 to become 6 hydrogen
Answer=3
First we have to find moles of C:
Molar mass of CO2:
12*1+16*2 = 44g/mol
(18.8 g CO2) / (44.00964 g CO2/mol) x (1 mol C/ 1 mol CO2) =0.427 mol C
Molar mass of H2O:
2*1+16 = 18g/mol
As there is 2 moles of H in H2O,
So,
<span>(6.75 g H2O) / (18.01532 g H2O/mol) x (2 mol H / 1 mol H2O) = 0.74mol H </span>
<span>Divide both number of moles by the smaller number of moles: </span>
<span>As Smaaler no moles is 0.427:
So,
Dividing both number os moles by 0.427 :
(0.427 mol C) / 0.427 = 1.000 </span>
<span>(0.74 mol H) / 0.427 = 1.733 </span>
<span>To achieve integer coefficients, multiply by 2, then round to the nearest whole numbers to find the empirical formula:
C = 1 * 2 = 2
H = 1.733 * 2 =3.466
So , the empirical formula is C2H3</span>
Answer:
Atoms
Explanation:
According to the law of conservation of matter, matter is neither created nor destroyed, so we must have the same number and kind of atoms after the chemical change as were present before the chemical change.
Answer:
See below ~
Explanation:
The calculated values of V/n :
⇒ 1.5/0.3 = 5
⇒ 3/0.6 = 5
⇒ 4.5/0.9 = 5
⇒ 6/1.2 = 5
⇒ 7.5/1.5 = 5
1. From this we understand that the calculated values of V/n remain constant, equal to <u>5</u> in this case.
2. The volume-mole graph will be a straight line passing through the origin. (Attached below)
The day's length is not a settled unit of time.
The earth itself is jiggling around space, going on an amazingly slight bended way and the unrest is dialing down record of tidal movement and the moon moving possibly away.
The day's length changes amid the time and in addition about predictably or something to that effect, and extra hop second is added to the timetable.
The second is portrayed using different unrests of the cesium 133 particle at 0 Kelvin in light of the fact that isolated from quantum mechanics, it isn't at risk to coincidental segments.
